1 mL of NaOH solution had been added This allowed us to calculate the molarity

1 ml of naoh solution had been added this allowed us

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NaOH solution had been added. This allowed us to calculate the molarity of the NaOH, which was .0109 M. For the second trial the solution reached the equivalence point after 31.8 mL, which calculates to the NaOH solution having a molarity of .0100 M. Both of our trials were extremely accurate because we used a .01 M solution of NaOH and using the equivalence points we were able to calculate molarities of .01 and .0109. Our trials were also precise because both of our calculated molarities are relatively close only differing by .0009. From the accuracy of our results I do not think that much error was present in this part of the experiment. However, error could of occurred in measuring outthe KHP or in accurately gauging how much of the NaOH solution has been released from the buret in the titration process.For Part B, we obtained the values of pH at different points in the dissociation, including the first halfway and equivalence point and the second halfway and equivalencepoint. A notable trend is that as more volume of NaOH is added, the pH of the solution increases because the solution is becoming more basic as the strong base, NaOH, is added. We also notice that near the equivalence points, the slope or the change in pH divided by the change in volume greatly increases. Our graph slowly increases with the addition of NaOH unitl the first equivalence point when the pH abruptly rises. Followingthis the pH continues to slowly increase with the addition of more NaOH. At the second equivalence point there is also an abrupt increase but it is much less intense than at the first equivalence point. This is because at equivalence points, the pH is very susceptible to the solution being added and will greatly change with just one drop to become more like the pH of the solution being added. By using these different pH values in the titration paired with their volumes, we were able to create a graph comparing the volumesand pH along with the 1stderivative of this plot. With these two data plots, we were able to visually determine where the equivalence points occurred and therefore we could calculate the halfway point. The first equivalence point took place at 6.5mL, meaning the halfway point, where pH is equal to pKa, took place at 6.5 mL. We could visually determine this by finding the first peak in the plot of the 1stderivative and by using that volume of the halfway point to obtain the pH on the first plot. The pH at the halfway point was 3.41 meaning that the pKa was also 3.41. From here, we can calculate the Ka value to be 3.89 x 10-4. This means that at the first equivalence point, the solution is still acidic and in this case, can deprotonate even more. To determine the second equivalence point, we find the second peak on the 1stderivative plot and use that volume to find the volume and pH (pKa) at the halfway point. The second equivalence point occurs at 19.25 mL. The pH (pKa) at this point is equal to 5.67, meaning the solution has become basic and the phosphoric acid has deprotonated again. We then calculate Ka to be 2.14 x 10-6. This Ka value is smaller than the value at the first equivalence point because less of the hydronium ion is being produced due to the added strong base, NaOH. H3PO4 is triprotic and should have 3 equivalence points, but the third point can’t be found graphically. The literature value for the first Kaa was 0.000750. The algebraic and graphical determination 
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  • Spring '11
  • Dr.JohnZhang
  • pH, Equivalence point, Sodium hydroxide, KHP

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